2009 AMC 10A Problems/Problem 1

Problem

One can hold $12$ ounces of soda. What is the minimum number of cans needed to provide a gallon ( $128$ ounces) of soda?

$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 9\qquad \mathrm{(D)}\ 10\qquad \mathrm{(E)}\ 11$

$10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\mathrm{(E)}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

We can divide $128/12$ and round up because there are a whole number of cans.

$128/12 = 10R8\longrightarrow 11\longrightarrow \fbox{E}.$